Reading Assignment: pp. 125-127 (start at bottom of p. 125), 133-134

31 Comments

Mick Piombino
on February 8, 2016 at 9:46 pm

I am confused about the int. value theorem, I need an example with numbers that are basic. Do you just always set the equation equal to 0 then solve? The limits that approach infinity and negative infinity made sense!

Jason Scott
on February 10, 2016 at 9:48 pm

Even after today’s class I’m still a little confused on the intermediate value theorem. Could you explain it a little more during class?

Jonathan E Lopez
on February 22, 2016 at 8:35 am

Hypothetical IVT example — suppose a person weighed 140 pounds at time 1, and 160 pounds at time 2 (let’s say f(1)=140, f(2)=160). Assuming that weight fluctuates continuously (you don’t “jump” from one weight to the next, but rather your weight gradually changes), I could ask for you to show me that at some time between times 1 and 2, the person weighted 150 pounds. In other words, show me there is a time value between 1 and 2 that satisfies f(t)=150. So show this using the IVT, one way to think of it is to consider the equation f(t)-150=0. If we can show this has a solution, then that means f(t)=150. This is equivalent to showing that the graph of y=f(t)-150 has a t-intercept. To show this (assuming we have already determined that y=f(t)-150 is continuous), we can just show that there is a point below the t-axis and a point above the t-axis. Since the function is continuous, the point below and the point above must be connected by an unbroken graph, which means the graph has to cross the t-axis at some point. This gives a t-value where f(t)-150=0, which means that at that t-value, f(t)=150 (i.e., the person’s weight was 150 at that time).

To show the point below and point above, you generally plug the endpoints of the interval into the function. In this case, note that

f(1)-150=140-150=-10 (so the point (1,-10) is on the graph); and
f(2)-150=160-150=10 (so the point (2,10) is on the graph).

This forces there to be at least one t-intercept between t=1 and t=2, which is a solution to our equation f(t)=150.

Anthony Rebmann
on February 11, 2016 at 11:52 am

Continuity is every confusing to me and I need to see examples up on the board that way I can grasp the concept of the theorem. It is hard to understand when they are using letters rather then numbers. I think it would be easier to understand a step by step in class with real numbers. Limitations on the other hand i think are a lot easier to understand because it is pretty straight forward.

Jonathan E Lopez
on February 22, 2016 at 8:38 am

We did a few continuity examples in class. In terms of a graph, a function is continous at a number x=a if the graph passes through the point (a,f(a)) without anything funny happening (a hole, a jump, a vertical asymptote). In order for this to happen, three things must occur:
1) f(a) must be a number (this means there is a point at (a,f(a));
2) lim_{x->a}{f(x)} must be a number (this means the left and right parts of the graph are headed to the same place near x=a); and
3) the number from 1) and the number from 2) must be the same (this means the place where the left and right pieces of the graph meet up is at the point (a,f(a)).

If all three of these things occur, then the function is continuous at x=a.

Samantha Caico
on February 11, 2016 at 7:04 pm

Are there other uses for the intermediate value theorem besides finding roots of equations? What if the function doesn’t cross the x axis at all, would the intermediate value theorem even matter; maybe it doesn’t cross due to a translation upward to a sin or cos function or doesn’t cross due to an asymptote. A graph could be continuous at some values and not others, right? Do the values of N have any significance or just a visual representation of the continuity of the graph within a certain interval? The method for solving infinite limits makes sense. But in example 4 it says for x<0, for the limit x approaches -infinity, that sqrt(x^2)=-x (which makes) but if they just divided the numerator by -x would that not work to solve for the horizontal asymptote as well, why or why not?

Jonathan E Lopez
on February 22, 2016 at 1:43 pm

With the IVT, we often translate the question into a question about roots. See the example above. We want to prove that at some time f(t)=150, where f(t)=weight at time t. To use the IVT, we work with the function y=f(t)-150, and a solution to f(t)=150 is the same as a root of y=f(t)-150. So although f(t) may not cross the t-axis, it’s possible that y=f(t)-150 does (and that’s what we try to show).

Yes, a function can definitely be continuous some places and not others. For example, f(x)=1/x is continuous at all numbers except x=0.

In the IVT, N represents the output of the function. The goal is to show that there is an x-value that gives output N (in other words, show that there is an x-value with f(x)=N). This is equivalent to showing the graph of y=f(x)-N crosses the x-axis. This is the idea used in the weight example described above.

In example 4, we need to divide sqrt(2x^2+1) by x. The problem is that we cannot bring x under the square root unless we rewrite it as a square root. This is why they say x = – sqrt(x^2). Here, since x–> – infty, note that x is negative. So if x = -100, then sqrt(x^2) = 100 = -x, which is the negative of what we started with. So for x<0, we have to use x = – sqrt(x^2). Once we know this, dividing by x is the same as dividing by x^2 under the radical (and inserting a minus sign for the reason described above). If we didn't put in the minus sign correctly, we would get the negative of the correct limit.

Alonda McEachin
on February 11, 2016 at 7:18 pm

On the concept of continuity, I understand the three rules which make a function continuous. Yesterday in class we went over functions that are not continuous and I a bit confused on how each function is categorized. What characteristic help us choose what type of discontinuity the function is? The IVT is also confusing, it basically says that a continuous function takes on every value between the function values f(a) and f(b), but how do we decide where these values are placed? Do limits play a role in IVT? On page 133, limits at infinity are discussed which is a relatively easy concept, as x gets closer to infinity or -infinity we can use the idea discussed in class of 1/big=small and 1/small=big. The limits at infinity also talk about dividing by the highest power found in the denominator. This also encompasses the 1/big=small and vice versa.

Jonathan E Lopez
on February 22, 2016 at 1:47 pm

Removable discontinuity means there is a hole in the graph (so either Q1=N and Q2=Y, or Q1=Q2=Y and Q3=N).

Jump discontinuity means Q2=N because the LHL and RHL both exist, but are different numbers.

Infinite discontinuity means Q2=N because some limit is infinity or -infinity.

When using the IVT, we generally know a “starting” value and an “ending” value (for example see the weight example described above). The IVT tells us that if the function is continuous, then every (intermediate) value between the starting and ending values is “hit”, i.e., is the output of the function for some input.

Limits aren’t directly part of the IVT, but they do play a role in continuity. Remember that the IVT can only be used if we know the function is continuous on the interval we are working with.

Jenna Page
on February 11, 2016 at 7:55 pm

For example 10 on page 126, they start with setting f(1) and f(2) which I understand, but at the end of each equation there is a 0 and I don’t know why those are there. For example 3 on page 133, they ask you to indicate which properties of limits are used at each stage. In the solution description it says you have to first divide both the numerators and denominators by the highest power of x in the denominator. How are you supposed to recognize when a problem requires you to do that?

Jonathan E Lopez
on February 22, 2016 at 1:51 pm

In example 10, their goal is to show there is a solution to f(x)=0, i.e., some input produces an output of 0.

They are noting that f(1)=-1 (the “starting value”) is negative and f(2)=12 (the “ending value”) is positive. Since the function is continuous, every value between the starting and ending values gets “hit”. Since 0 is between -1 and 12, it gets “hit”, which means there is an x-value that makes f(x)=0.

The limit properties are basically the fact that we can look at the limit of each “piece”, as we have done in class.

The technique of dividing by the highest power of x only works for limits as x–> infty or -infty. Also, the function would have to basically look like a polynomial over a polynomial, or possibly with a square root like example 4 on page 134.

Alissa Buchta
on February 11, 2016 at 8:21 pm

I understand the ideas of continuity and the idea that in between y=f(a) and y=f(b) will be y=N however example 10 confuses me. How they came to the conclusion that the root must lie between 1.2 and 1.3 is confusing to me because I’m not really understanding where the numbers -0.128 and 0.548 are coming from. I was able to understand example 3 on page 133 because I was able to follow it based on the examples we did in class today although it was a little confusing due to the fact that there were several limits and algebraic equations involved.

Jonathan E Lopez
on February 22, 2016 at 1:56 pm

They are sorting repeating the use of the IVT to improve their estimate. Initially, they show that since f(1)=-1 and f(2)=12, there must be an x-value between x=1 and x=2 that makes f(x)=0. But this is not a very precise estimate of the actual solution. Note that the idea was to get one point above the x-axis and one point below the x-axis. Their reasoning is that if they can find other x-values with these properties that are closer together, then the estimate for the solution will be better.

So they basically pick x=1.2 and plug into the function f to get -0.128. They pick x=1.3 and plug into the function to get 0.548. So again, we have one point above the x-axis and one point below. So there must be place in between where the graph crosses the x-axis. So this tells us that there is a solution of f(x)=0 between x=1.2 and x=1.3. This is much more precise than just saying the solution is between x=1 and x=2. The way they picked x=1.2 and x=1.3 is likely by trial and error (again, they are looking for one point above and one point below the x-axis).

Emily Czechowski
on February 11, 2016 at 9:00 pm

The wording of the intermediate value theorem in the textbook was rather confusing. The graphs helped illustrate the theorem better and as I read on it got more clear. However it is still kind of confusing…is it basically just stating that if a function is continuous then there has to be a value somewhere between 2 values of the function? The information about the limits at infinity and horizontal asymptotes makes more sense to me. The graphs again helped me better understand this along with the examples they provided. If you can use your calculator or if a picture of a graph is given to you can you just give an answer based off of that or would you have to solve it algebraically still?

Jonathan E Lopez
on February 22, 2016 at 1:59 pm

Yes, think of having a starting and ending value. If the function is continuous, then every value in between those is “hit”.

Think of a population of bacteria where fractional bacteria are possible. If we know that at t=0, there are 100 bacteria and at time t=5, there are 1,000 bacteria, then at some point in time (between t=0 and t=5), there were 200 bacteria, and at some time there were 500 bacteria, and at some time there were 785.43 bacteria, etc.

If the instructions say to find the limit “algebraically”, it means to use the techniques we’ve been using in class. If not, then the calculator or the graph can be helpful.

Jenivette Garcia
on February 11, 2016 at 9:53 pm

The intermediate value theorem’s definition makes sense. It makes sense because if there is a y coordinate between 2 y coordinates, and the function is smooth, theres obviously an x coordinate for that y coordinate. IVT becomes confusing when you try using it on an equation..
Horizontal asymptotes, on the other hand are easy to find because if the degree of the numerator and denominator are the same, then the limit is the ratio of the leading coefficients. If the degrees are different then the limit is 0.

Jonathan E Lopez
on February 22, 2016 at 2:00 pm

If the degree of the denominator is bigger, then limit = 0. But if the degree of the numerator is bigger, then limit is either infinity or -infinity.

Melanie Rivera
on February 11, 2016 at 10:17 pm

I understand the Intermediate Value Theorem and how f(c) must equal N but the fifth theorem on page 133 is a bit confusing to me. I do not quite understand what the second part of the theorem is trying to convey. I would like a bit more clarification.

Jonathan E Lopez
on February 22, 2016 at 2:02 pm

The second part of Theorem 5 is basically saying that we must be careful of using negative x-values with even roots.

If r=1/2, for example, then the function is f(x)=1/sqrt(x). This function does not make sense for negative x-values, so the limit of this function as x–> -infinity is DNE (in particular, it’s not 0).

So basically when x–> -infinity, as long as the function 1/x^r makes sense for negative x-values, then the limit is 0.

Kianne Fernandez
on February 11, 2016 at 10:31 pm

The Intermediate Value Theorem basically states that there aren’t any holes or jumps in the graph at all. There has to be a value between two numbers or points on the graph. I just don’t understand how you prove it. I understand that there are roots but I can’t remember how to find the roots? I remember there were rules for the limits of infinity which made it easy to figure out what the answer is. There is also another way in the calculator that makes it easy to answer. Go to the table set, put 1,000,000 in the table start and below that in the next line, put .0001. Then go to the table in the calculator and the y-values will give you the answer.

Jonathan E Lopez
on February 22, 2016 at 2:06 pm

See the example with the weight function described above. Note that the IVT does not tell us that a function is continuous. IF the function is continuous, THEN we are allowed to use the IVT.

As far as finding the roots, we generally aren’t doing that when using the IVT (usually because finding the roots is difficult or impossible). So instead, we just show that there IS a root (without finding its value). This is equivalent to showing that the graph has an x-intercept.

Tim Ler
on February 11, 2016 at 10:39 pm

For reading 125-127 the only thing I’m confused about is discontinuities and the theorems. Are discontinuities breaks in the graph? And if it’s continuous the domain has to be all real numbers? And for 133-134 theorem 5 says ” if r > 0 is a rational number such that x^r is defined for all x” what does that mean? And do you really have to do all this simplyfing? Can’t you just take the leading coefficients with the highest matching powers? 🤔

Jonathan E Lopez
on February 22, 2016 at 2:10 pm

Yes, discontinuities would be some sort of problem with the graph. It could be a hole, a jump, a vertical asymptote, or the complete lack of a graph.

If a function is continuous, it’s domain might not be all real numbers. For example, we say f(x)=sqrt(x) is continous on it’s domain, but it’s not continuous at x=-1, for example.

For theorem 5, the second part is saying that if x is negative, we have to be careful if doing even roots. There is a comment above about this. If r=1/2, for example, the limit as x–> -infty of 1/sqrt(x) is not 0, because 1/sqrt(x) is not defined for negative x-values.

If the numerator and denominator have the same degree, then the limit should be the ratio of leading coefficients. The technique of dividing by the highest power of x in the denominator is how we justify that answer. In this course, the answer is less important than the method in which the answer is obtained.

Peter Bartnik
on February 11, 2016 at 10:42 pm

For the intermediate value Theorem, is example 10 an example of the IVT being used to show that a solution is possible? It doesn’t specifically ask if a solution is possible, but that seems to be the conclusion since a definitive answer isn’t given. If that is the case, it would have been nice if the book showed a case where the IVT could be used to prove that a solution isn’t possible.

Jonathan E Lopez
on February 22, 2016 at 2:13 pm

Yes, they say “show there is a root”, which really means “show there is a solution”. Think of “root” and “solution” as synonyms.

Unfortunately, the IVT cannot by itself prove that a solution isn’t possible. If we also included something about the function being differentiable and never having f'(x)=0 on the interval, then these two ideas together could be used to show that an equation does not have a solution (but that is beyond the scope of this course).

Therese Hetzer
on February 11, 2016 at 10:59 pm

For this reading 133-134 was very understandable. Both example 3 and 4 were easy for me to follow along and the in class work we did on the topic of limits at infinity and horizontal asymptotes was clear. I do struggle with continuity and the IVT when it comes to the actual equation. The problem (s) done in class made sense with the step by step directions, but looking at example 9 and 10 I am confused. Example 9 may just require me to brush up more on trig functions, but 10 confuses me when it comes to getting down to the specific values/roots. “A root lies between (1.22, 1.23)” How? Do we use the calculator to find them, since that seems to be how to do it in the book? Is there a way to do it without the calculator? Maybe the layout of the explanation/solution is throwing me off.

Jonathan E Lopez
on February 22, 2016 at 2:16 pm

See the comment above regarding this example. Typically, we would stop once we have shown there is a solution between x=1 and x=2. But they go on in that example to try to improve the estimate of the solution by finding two x-values that are closer together (and which still satisfy one point above and one point below the x-axis).

Since f(1) is negative and we are interested in where f(x)=0, think of it as trying to figure out where f changes from negative to positive. So we know f(1) is negative, f(1.1) is still negative, f(1.2) is still negative, but f(1.3) is positive. So somewhere between x=1.2 and x=1.3, the graph crosses the x-axis.

Marnisha Brooks
on February 11, 2016 at 11:01 pm

On pages 125-127 I understand that the intermediate value theorem is a continuous function and are the numbers between the given interval but when the book went into showing that there was a root for that equation that completely threw me off. I don’t understand how they got f(2) at all. But on pages 133-134 I understand the concept of limits at infinity and finding the vertical and horizontal asymptotes. Us doing examples in class helped me understand the reading and examples in the book.

Jonathan E Lopez
on February 22, 2016 at 2:17 pm

To use the IVT, we need a continuous function on an interval (like (1,2), for example). The IVT tells us that every y-value between f(1) and f(2) is “hit” by some x-value.

To get f(2), plug x=2 into the equation for f(x):

f(2)=4(2^3)-6(x^2)+3(2)-2
=4(8)-6(4)+6-2
=32-24+6-2
=8+6-2
=14-2
=12.

Dan Emerson
on February 11, 2016 at 11:52 pm

The intermediate value theorem is very confusing. The graphs make it more confusing, but I get an understanding from the written definition. I do not see the relationship when looking at the graphs however. The asymptotes are something I am familiar with, so I understood those.

Jonathan E Lopez
on February 22, 2016 at 2:21 pm

In terms of graphs, if f is continuous on an interval (a,b), then every y-value between f(a) and f(b) is the y-value of a point on the graph (so if you pick any number N between f(a) and f(b), the line y=N should intersect the graph of f at least once).

Rather than showing there is an x-value with f(x)=N as described above, it’s common to work with f(x)-N=0 instead, and showing that there is an x-value that satisfies this equation is the same as showing that the graph of y=f(x)-N has an x-intercept. If f is continuous, then showing there is an x-intercept just requires showing there is a point on the graph below the x-axis and a point on the graph above the x-axis (generally f(a) and f(b) are used, and one is below and one is above).